A355641 -id:A355641 - cp's OEIS frontend

A355200 Numbers k that can be written as the sum of 3 divisors of k (not necessarily distinct).

3, 4, 6, 8, 9, 12, 15, 16, 18, 20, 21, 24, 27, 28, 30, 32, 33, 36, 39, 40, 42, 44, 45, 48, 51, 52, 54, 56, 57, 60, 63, 64, 66, 68, 69, 72, 75, 76, 78, 80, 81, 84, 87, 88, 90, 92, 93, 96, 99, 100, 102, 104, 105, 108, 111, 112, 114, 116, 117, 120, 123, 124, 126, 128, 129, 132, 135

Offset: 1

Wesley Ivan Hurt, Jun 23 2022

From Bernard Schott, Aug 06 2023: (Start)
Equivalently: positive numbers that are divisible by 3 or by 4.
Proof (similar to the proof proposed by Robert Israel in A355641).
If k is divisible by 3, then k is in the sequence because k = k/3 + k/3 + k/3.
If k is divisible by 4, then k is in the sequence because k = k/2 + k/4 + k/4.
Moreover, if k is positive and divisible by 6 (A008588), then k = k/3 + k/3 + k/3, but k is also in the sequence because k = k/2 + k/3 + k/6.
Conversely, to show that every term of this sequence is divisible by 3 or by 4, we consider all positive integer solutions of the equation 1 = 1/a + 1/b + 1/c. Without loss of generality, we may assume a <= b <= c, then 3/a >= 1/a + 1/b + 1/c = 1. So a <= 3. Similarly, given a, we have 2/b >= 1/b + 1/c = 1 - 1/a, so b <= 2/(1 - 1/a).
-> if a = 1, then 1 = 1 + 1/b + 1/c; this equation has clearly no solution.
-> if a = 2, then 1/2 = 1/b + 1/c with b <= 2/(1 - 1/2) = 4; in this case, there are two solutions: (a,b,c) = (2,3,6) or (a,b,c) = (2,4,4).
-> if a = 3, then 2/3 = 1/b + 1/c with b <= 2/(1 - 1/3) = 3; in this case, there is one solution: (a,b,c) = (3,3,3).
It turns out that there are only 3 solutions with a <= b <= c. Each corresponds to a possible pattern k = k/a + k/b + k/c for writing k as the sum of 3 of its divisors, which works when k is divisible by 3 or by 4. (End)
From David A. Corneth, Aug 07 2023: (Start)
Proof that a(n + 6) = a(n) + 12.
As k is in the sequence, k = k/d1 + k/d2 + k/d3 where d1, d2 and d3 | k and they are not necessarily distinct. By discussion above from Bernard Schott, Aug 06 2023, (d1, d2, d3) are in {(2, 3, 6), (2, 4, 4), (3, 3, 3)}. The lcm of these tuples are 6, 4 and 3 respectively. So any number k in the sequence is divisible by 3, 4 or 6.
This tells us that if k is in the sequence then k + 12 is in the sequence since k + 12 is divisible by one of 3, 4 or 6 since lcm(3, 4, 6) = 12.
So we can write a(n + m) = a(n) + 12 for some m. Inspection gives m = 6 so a(n + 6) = a(n) + 12. (End)

			6 is in the sequence since it can be written as the sum of 3 of its (not necessarily distinct) divisors: 6 = 1+2+3 = 2+2+2 with 1|6, 2|6, and 3|6.

Ray Chandler, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (2,-2,2,-2,2,-1).

Cf. A002966, A008588, A348536.
Numbers k that can be written as the sum of j divisors of k (not necessarily distinct) for j=1..10: A000027 (j=1), A299174 (j=2), this sequence (j=3), A354591 (j=4), A355641 (j=5), A356609 (j=6), A356635 (j=7), A356657 (j=8), A356659 (j=9), A356660 (j=10).
Equals positive terms of A008585 union A008586.
Cf. A005843, A134667.

q[n_, k_] := AnyTrue[Tuples[Divisors[n], k], Total[#] == n &]; Select[Range[135], q[#, 3] &] (* Amiram Eldar, Aug 21 2022 *)
Table[2n + (Sin[Pi*n/3] + Sin[2*Pi*n/3])/Sqrt[3], {n, 100}] (* Wesley Ivan Hurt, Oct 30 2023 *)
CoefficientList[Series[(3 - 2*x + 4*x^2 - 2*x^3 + 3*x^4)/((x - 1)^2*(1 + x^2 + x^4)), {x, 0, 80}], x] (* Wesley Ivan Hurt, Jul 17 2025 *)

isok(k) = my(d=divisors(k)); forpart(p=k, if (setintersect(d, Set(p)) == Set(p), return(1)), , [3,3]); \\ Michel Marcus, Aug 21 2022

is(n) = my(v = [3,4,6]); sum(i = 1, 3, n%v[i] == 0) > 0 \\ David A. Corneth, Oct 08 2022

a(n + 6) = a(n) + 12. - David A. Corneth, Oct 08 2022
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*log(2)/3 - log(3)/4. - Amiram Eldar, Sep 10 2023
From Wesley Ivan Hurt, Oct 30 2023: (Start)
a(n) = 2*n + (sin(Pi*n/3) + sin(2*Pi*n/3))/sqrt(3).
a(n) = A005843(n) + A134667(n). (End)
From Wesley Ivan Hurt, Jul 17 2025: (Start)
G.f.: x*(3-2*x+4*x^2-2*x^3+3*x^4)/((x-1)^2*(1+x^2+x^4)).
a(n) = 2*a(n-1) - 2*a(n-2) + 2*a(n-3) - 2*a(n-4) + 2*a(n-5) - a(n-6). (End)

A354591 Numbers k that can be written as the sum of 4 divisors of k (not necessarily distinct).

Original entry on oeis.org

4, 6, 8, 10, 12, 16, 18, 20, 24, 28, 30, 32, 36, 40, 42, 44, 48, 50, 52, 54, 56, 60, 64, 66, 68, 70, 72, 76, 78, 80, 84, 88, 90, 92, 96, 100, 102, 104, 108, 110, 112, 114, 116, 120, 124, 126, 128, 130, 132, 136, 138, 140, 144, 148, 150, 152, 156, 160, 162, 164, 168, 170, 172

Offset: 1

Wesley Ivan Hurt, Aug 18 2022

nonn

All terms are even. - Robert Israel, Aug 31 2022
Is it true that a(n) = 2*A080671(n)? - Michel Marcus, Sep 01 2022 (True for n <= 10000. - N. J. A. Sloane, Sep 01 2022)
This is true.  In other words, k is in the sequence if and only if k is even and divisible by 3, 4 or 5.  Proof: the positive integer solutions of 1/a + 1/b + 1/c + 1/d = 1 can be enumerated explicitly, and each contains at least one even number and at least one divisible by 3, 4 or 5.  Of course k = k/a + k/b + k/c + k/d if and only if 1 = 1/a + 1/b + 1/c + 1/d, and this writes k as the sum of 4 divisors of k if k is divisible by a,b,c, and d.  If k is even and divisible by 3, we can use 1 = 1/3 + 1/3 + 1/6 + 1/6; if divisible by 4, 1 = 1/4 + 1/4 + 1/4 + 1/4; if even and divisible by 5, 1 = 1/2 + 1/5 + 1/5 + 1/10. - Robert Israel, Sep 01 2022
The asymptotic density of this sequence is 11/30. - Amiram Eldar, Aug 08 2023

			20 is in the sequence since 20 = 10+5+4+1 = 5+5+5+5 where each summand divides 20.

Robert Israel, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (2,-2,2,-2,2,-2,2,-2,2,-2,2,-2,2,-2,2,-2,2,-2,2,-2,2,-1).

Numbers k that can be written as the sum of j divisors of k (not necessarily distinct) for j=1..10: A000027 (j=1), A299174 (j=2), A355200 (j=3), this sequence (j=4), A355641 (j=5), A356609 (j=6), A356635 (j=7), A356657 (j=8), A356659 (j=9), A356660 (j=10).

Cf. A080671.

F:= proc(x,S,j) option remember;
      local s,k;
      if j = 0  then return(x = 0) fi;
      if S = [] or x > j*S[-1] then return false fi;
      s:= S[-1];
      for k from 0 to min(j,floor(x/s)) do
        if procname(x-k*s, S[1..-2],j-k) then return true fi
      od;
      false
end proc:
select(t -> F(t, sort(convert(numtheory:-divisors(t),list)),4), [$1..200]); # Robert Israel, Aug 31 2022

q[n_, k_] := AnyTrue[Tuples[Divisors[n], k], Total[#] == n &]; Select[Range[200], q[#, 4] &] (* Amiram Eldar, Aug 19 2022 *)
CoefficientList[Series[2 (2 - x + 2*x^2 - x^3 + 2*x^4 + x^6 + 2*x^8 + x^10 + 2*x^12 + x^14 + 2*x^16 - x^17 + 2*x^18 - x^19 + 2*x^20)/((x^10 - x^9 + x^8 - x^7 + x^6 - x^5 + x^4 - x^3 + x^2 - x + 1)*(1 + x + x^5 + x^6 + x^7 + x^8 + x^9 + x^2 + x^4 + x^3 + x^10)*(x - 1)^2), {x, 0, 50}], x] (* Wesley Ivan Hurt, Jul 17 2025 *)

isok(k) = my(d=divisors(k)); forpart(p=k, if (setintersect(d, Set(p)) == Set(p), return(1)), , [4,4]); \\ Michel Marcus, Aug 19 2022

a(n) = 2*a(n-1) - 2*a(n-2) + 2*a(n-3) - 2*a(n-4) + 2*a(n-5) - 2*a(n-6) + 2*a(n-7) - 2*a(n-8) + 2*a(n-9) - 2*a(n-10) + 2*a(n-11) - 2*a(n-12) + 2*a(n-13) - 2*a(n-14) + 2*a(n-15) - 2*a(n-16) + 2*a(n-17) - 2*a(n-18) + 2*a(n-19) - 2*a(n-20) + 2*a(n-21) - a(n-22). - Wesley Ivan Hurt, Jun 29 2024

G.f.: 2*x*(2 - x + 2*x^2 - x^3 + 2*x^4 + x^6 + 2*x^8 + x^10 + 2*x^12 + x^14 + 2*x^16 - x^17 + 2*x^18 - x^19 + 2*x^20)/((x^10 - x^9 + x^8 - x^7 + x^6 - x^5 + x^4 - x^3 + x^2 - x + 1)*(1 + x + x^5 + x^6 + x^7 + x^8 + x^9 + x^2 + x^4 + x^3 + x^10)*(x - 1)^2). - Wesley Ivan Hurt, Jul 17 2025

A356609 Numbers k that can be written as the sum of 6 divisors of k (not necessarily distinct).

Original entry on oeis.org

6, 8, 10, 12, 14, 16, 18, 20, 24, 28, 30, 32, 36, 40, 42, 44, 48, 50, 52, 54, 56, 60, 64, 66, 70, 72, 78, 80, 84, 88, 90, 96, 98, 100, 102, 104, 108, 110, 112, 114, 120, 126, 128, 130, 132, 136, 138, 140, 144, 150, 152, 154, 156, 160, 162, 168, 170, 174, 176, 180, 182, 184, 186, 190

Offset: 1

Wesley Ivan Hurt, Aug 18 2022

nonn

Numbers divisible by at least one of 6, 8, 10, 14, 44, 52. For proof see link. - Robert Israel, Sep 02 2022

The asymptotic density of this sequence is 483/1430. - Amiram Eldar, Aug 08 2023

			18 is in the sequence since 18 = 9+2+2+2+2+1, where each summand divides 18.

Robert Israel, Table of n, a(n) for n = 1..10000
Robert Israel, Proof that A356609 consists of all numbers divisible by at least one of 6, 8, 10, 14, 44, 52.

Numbers k that can be written as the sum of j divisors of k (not necessarily distinct) for j=1..10: A000027 (j=1), A299174 (j=2), A355200 (j=3), A354591 (j=4), A355641 (j=5), this sequence (j=6), A356635 (j=7), A356657 (j=8), A356659 (j=9), A356660 (j=10).

filter:= n-> ormap(t -> n mod t = 0, [6,8,10,14,44,52]):
select(filter, [$1..200]); # Robert Israel, Sep 02 2022

q[n_, k_] := AnyTrue[Tuples[Divisors[n], k], Total[#] == n &]; Select[Range[200], q[#, 6] &] (* Amiram Eldar, Aug 19 2022 *)

isok(k) = my(d=divisors(k)); forpart(p=k, if (setintersect(d, Set(p)) == Set(p), return(1)), , [6,6]); \\ Michel Marcus, Aug 19 2022

A356635 Numbers k that can be written as the sum of 7 divisors of k (not necessarily distinct).

Original entry on oeis.org

7, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 27, 28, 30, 32, 33, 35, 36, 39, 40, 42, 44, 45, 48, 49, 50, 52, 54, 55, 56, 60, 63, 64, 66, 68, 70, 72, 75, 77, 78, 80, 81, 84, 88, 90, 91, 96, 98, 99, 100, 102, 104, 105, 108, 110, 112, 114, 117, 119, 120, 126, 128, 130

Offset: 1

Wesley Ivan Hurt, Aug 18 2022

nonn

If k is in the sequence then so is k*m for positive m. - David A. Corneth, Aug 19 2022
Numbers that are divisible by at least one of 7, 8, 9, 10, 12, 15, 22, 33, 39, 52, 55, 68, 102, 114, 138.  For proof, see link. - Robert Israel, Sep 02 2022
The asymptotic density of this sequence is 17819629/37182145 = 0.479252... . - Amiram Eldar, Aug 08 2023

			10 is in the sequence since 10 = 2+2+2+1+1+1+1, where each summand divides 10.

Robert Israel, Table of n, a(n) for n = 1..10000
Robert Israel, Proof that A356635 consists of all numbers divisible by at least one of 7, 8, 9, 10, 12, 15, 22, 33, 39, 52, 55, 68, 102, 114, 138.

Numbers k that can be written as the sum of j divisors of k (not necessarily distinct) for j=1..10: A000027 (j=1), A299174 (j=2), A355200 (j=3), A354591 (j=4), A355641 (j=5), A356609 (j=6), this sequence (j=7), A356657 (j=8), A356659 (j=9), A356660 (j=10).

filter:= n -> ormap(t -> n mod t = 0, [7, 8, 9, 10, 12, 15, 22, 33, 39, 52, 55, 68, 102, 114, 138]):
select(filter, [$1..200]); # Robert Israel, Sep 02 2022

q[n_, k_] := AnyTrue[Tuples[Divisors[n], k], Total[#] == n &]; Select[Range[130], q[#, 7] &] (* Amiram Eldar, Aug 19 2022 *)

isok(k) = my(d=divisors(k)); forpart(p=k, if (setintersect(d, Set(p)) == Set(p), return(1)), , [7,7]); \\ Michel Marcus, Aug 19 2022

A356657 Numbers k that can be written as the sum of 8 divisors of k (not necessarily distinct).

Original entry on oeis.org

8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 36, 40, 42, 44, 48, 50, 52, 54, 56, 60, 64, 66, 68, 70, 72, 76, 78, 80, 84, 88, 90, 96, 98, 100, 102, 104, 108, 110, 112, 114, 120, 126, 128, 130, 132, 136, 138, 140, 144, 150, 152, 154, 156, 160, 162, 168, 170, 174, 176

Offset: 1

Wesley Ivan Hurt, Aug 20 2022

nonn

Terms are even. Proof by contradiction. Suppose m = a(n) is odd. Then each divisor is odd. Adding 8 odd numbers gives an even number. A contradiction. - David A. Corneth, Sep 02 2022

			14 is in the sequence since 14 = 2+2+2+2+2+2+1+1, where each summand divides 14.

David A. Corneth, Table of n, a(n) for n = 1..10000

Numbers k that can be written as the sum of j divisors of k (not necessarily distinct) for j=1..10: A000027 (j=1), A299174 (j=2), A355200 (j=3), A354591 (j=4), A355641 (j=5), A356609 (j=6), A356635 (j=7), this sequence (j=8), A356659 (j=9), A356660 (j=10).

isok(k) = my(d=divisors(k)); forpart(p=k, if (setintersect(d, Set(p)) == Set(p), return(1)), , [8,8]); \\ Michel Marcus, Aug 21 2022

is(n) = if(n % 2 == 1, return(0)); my(d = divisors(n)); forvec(x = vector(8, i, [1, #d-1]), s=sum(i=1, #x, d [x[i]]); if(n == s, print(vector(#x, j, d[x[j]]));return(1)), 1); 0 \\ David A. Corneth, Aug 21 2022

A356659 Numbers k that can be written as the sum of 9 divisors of k (not necessarily distinct).

Original entry on oeis.org

9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 35, 36, 39, 40, 42, 44, 45, 48, 50, 51, 52, 54, 55, 56, 57, 60, 63, 64, 65, 66, 68, 70, 72, 75, 76, 77, 78, 80, 81, 84, 85, 88, 90, 92, 96, 98, 99, 100, 102, 104, 105, 108, 110, 112, 114, 117, 120, 125

Offset: 1

Wesley Ivan Hurt, Aug 20 2022

nonn

If k is in the sequence then so is k*m. - David A. Corneth, Oct 08 2022

			14 is in the sequence since 14 = 2+2+2+2+2+1+1+1+1, where each summand divides 14.

David A. Corneth, Table of n, a(n) for n = 1..10000

Numbers k that can be written as the sum of j divisors of k (not necessarily distinct) for j=1..10: A000027 (j=1), A299174 (j=2), A355200 (j=3), A354591 (j=4), A355641 (j=5), A356609 (j=6), A356635 (j=7), A356657 (j=8), this sequence (j=9), A356660 (j=10).

upto(n) = { my(v = vector(n,i,-1), t = 0); for(i = 1, n, if(v[i] == -1, print1(i", "); v[i] = is(i, 9); if(v[i] == 1, for(j = 2, n \ i, v[i*j] = 1; ) ) ); ); select(x->x >= 1, v, 1); }
is(n, {qd = 10}) = { my(d = divisors(n)); d = d[^#d]; forvec(x = vector(qd-1, i, [1, #d]), s = sum(i = 1, qd-1, d[x[i]]); if(n - s >= d[x[qd - 1]], if(n % (n - s) == 0, return(1); ) ) , 1 ); 0 } \\ David A. Corneth, Oct 08 2022

a(n + t) = a(n) + s for some finite t and s. - David A. Corneth, Oct 08 2022

A356660 Numbers k that can be written as the sum of 10 divisors of k (not necessarily distinct).

Original entry on oeis.org

10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 40, 42, 44, 48, 50, 52, 54, 56, 60, 64, 66, 68, 70, 72, 76, 78, 80, 84, 88, 90, 92, 96, 98, 100, 102, 104, 108, 110, 112, 114, 116, 120, 126, 128, 130, 132, 136, 138, 140, 144, 150, 152, 154, 156, 160, 162

Offset: 1

Wesley Ivan Hurt, Aug 20 2022

nonn

From David A. Corneth, Oct 08 2022: (Start)
All terms are even. Proof: suppose a term is odd. Then all divisors are odd. Adding 10 odd numbers gives an even number. A contradiction.
If k is a term then so is k*m for m >= 1. Proof: Multiply each divisor in this sum of 10 divisors that give k with m. Then each term is a divisor of k*m and their sum is k*m. (End)

			14 is in the sequence since 14 = 2+2+2+2+1+1+1+1+1+1, where each summand divides 14.

David A. Corneth, Table of n, a(n) for n = 1..10000

Numbers k that can be written as the sum of j divisors of k (not necessarily distinct) for j=1..10: A000027 (j=1), A299174 (j=2), A355200 (j=3), A354591 (j=4), A355641 (j=5), A356609 (j=6), A356635 (j=7), A356657 (j=8), A356659 (j=9), this sequence (j=10).

upto(n) = { my(v = vector(n,i,-1), t = 0); forstep(i = 2, n, 2, if(v[i] == -1, v[i] = is(i); if(v[i] == 1, for(j = 2, n \ i, v[i*j] = 1; ) ) ); ); select(x->x >= 1, v, 1); }
is(n, {qd = 10}) = { my(d = divisors(n), res = 0); d = d[^#d]; forvec(x = vector(qd-1, i, [1, #d]), s = sum(i = 1, qd-1, d[x[i]]); if(n - s >= d[x[qd - 1]], if(n % (n - s) == 0,  return(1); ) ) , 1 ); 0 } \\ David A. Corneth, Oct 08 2022

from sympy import divisors
def t_sum_of_n_div(n, target):
    out, p = [], divisors(n)[::-1][1:]
    def dfs(t, divs,  index_s, kk):
        if len(out)!=0 or kk>target:return
        if kk == target and t == 0:
            out.append(divs)
            return
        for i in range(index_s, len(p)):
            if t >= p[i]:
                temp_divs = divs.copy()
                temp_divs.append(p[i])
                dfs(t-p[i], temp_divs, i, kk+1)
    dfs(n, [], 0, 0)
    return out
terms = [i for i in range(2, 200) if len(t_sum_of_n_div(i,10))!=0]
print(terms) # Gleb Ivanov, Sep 02 2022

A356919 Number of partitions of n into 5 parts that divide n.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 0, 2, 1, 2, 0, 8, 0, 1, 3, 3, 0, 6, 0, 6, 1, 0, 0, 20, 1, 0, 1, 2, 0, 14, 0, 3, 0, 0, 1, 20, 0, 0, 0, 11, 0, 8, 0, 0, 5, 0, 0, 26, 0, 2, 0, 0, 0, 7, 1, 4, 0, 0, 0, 41, 0, 0, 2, 3, 1, 2, 0, 0, 0, 5, 0, 35, 0, 0, 3, 0, 0, 2, 0, 12, 1, 0, 0, 25, 1, 0, 0, 2, 0, 23, 0, 0, 0, 0, 1, 27, 0, 1, 1, 7, 0, 1, 0, 2, 4

Offset: 1

Wesley Ivan Hurt, Sep 04 2022

nonn

			a(12) = 8; there are 8 ways to write 12 as the sum of 5 divisors of 12: 6+3+1+1+1 = 6+2+2+1+1 = 4+4+2+1+1 = 4+3+3+1+1 = 4+3+2+2+1 = 4+2+2+2+2 = 3+3+3+2+1 = 3+3+2+2+2.

Antti Karttunen, Table of n, a(n) for n = 1..20000
Index entries for sequences related to partitions

Cf. A355641.

A356919(n, x=n, y=n, nparts=5) = if(0==nparts, (0==y), if(y<=0, 0, sumdiv(n, d, if((d<=x), A356919(n, d, y-d, nparts-1))))); \\ Antti Karttunen, Jan 13 2025

a(n) = Sum_{l=1..floor(n/5)} Sum_{k=l..floor((n-l)/4)} Sum_{j=k..floor((n-k-l)/3)} Sum_{i=j..floor((n-j-k-l)/2)} c(n/l) * c(n/k) * c(n/j) * c(n/i) * c(n/(n-i-j-k-l)), where c(n) = 1 - ceiling(n) + floor(n).

More terms from Antti Karttunen, Jan 13 2025

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A355200 Numbers k that can be written as the sum of 3 divisors of k (not necessarily distinct).

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A354591 Numbers k that can be written as the sum of 4 divisors of k (not necessarily distinct).

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A356609 Numbers k that can be written as the sum of 6 divisors of k (not necessarily distinct).

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A356635 Numbers k that can be written as the sum of 7 divisors of k (not necessarily distinct).

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A356657 Numbers k that can be written as the sum of 8 divisors of k (not necessarily distinct).

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A356659 Numbers k that can be written as the sum of 9 divisors of k (not necessarily distinct).

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A356660 Numbers k that can be written as the sum of 10 divisors of k (not necessarily distinct).

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